Geometric Sequence

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A geometric sequence is a sequence whose terms change by a constant factor.

A few examples of sequences are below:

(1)$5,25,125,625,3125\,$

(2)$81,-27,9,-3,1,-\frac{1}{3}$

It should be apparent that the two sequences are increasing by 5 and $-\frac{1}{3}$, respectively. As a result, we can determine the value of any given term in the sequence if we know its position.

We examine sequence (1) to find this formula

We identify the first term, 5, as $a_0$ and the following terms $25, 125, 625,... \,$ as $a_1, a_2, a_3,...\,$.

We observe

$a_1 = a_0 * 5\,$

$a_2 = a_1 * 5 = a_0 * 5^2\,$

$a_3 = a_2 * 5 = a_0 * 5^3\,$

$a_4 = a_3 * 5 = a_0 * 5^4\,$

From here we can reason that $a_n = a_0 * 5^n\,$.

We can generalize further that the constant factor of change does not need to be 5 but actually can be any value r.

Thus we have the expression

$a_n = a_0 * r^n\,$

At this point, we must make the distinction between the words series and sequence. A sequence is a list of numbers of finite or infinite length; for example, (1) is geometric sequence of length 5. A series is a sum of numbers of a sequence.

Other Properties

Another property of geometric series is that any term is the square root of the product of its two neighbors. For example:

$\sqrt{5*125} = \sqrt{5^4} = 5^2\ =25,$

In fact, so long as the absolute difference between the two values to be square rooted is the same integer, the property will hold.

$a_n = a_0 * 5^n = \sqrt{(a_0 * 5^{(n-d)})*(a_0 * 5^{(n+d)})}$

Summing the Sequence

A very convenient formula for summing a geometric sequence can found from algebraic manipulation.

$s = a + ar + ar^2 + ar^3 + \cdots$

We divide both sides by a,

(1) $\frac{s}{a} = 1 + r + r^2 + r^3 + \cdots$

Here we multiply both sides by r,

(2) $\frac{rs}{a} = r + r^2 + r^3 + r^4 + \cdots$

We now subtract equation (2) from (1), which results in

$\frac{s - rs}{a} = 1$

Simplifying for s gives us,

$s = \frac{a}{1-r}$

This gives us the formula for an infinite geometric series.

Similarly, we use a similar process to find the sum of a finite geometric series.

$s = a + ar + ar^2 + ar^3 + \cdots + ar^n$

We divide both sides by a,

(3)$\frac{s}{a} = 1 + r + r^2 + r^3 + \cdots + r^n$

Here we multiply both sides by r,

(4)$\frac{rs}{a} = r + r^2 + r^3 + r^4 + \cdots + r^{n+1}$

We now subtract equation (4) from (3), which results in

$\frac{s - rs}{a} = 1 - r^{n+1}$

Simplifying for s gives us,

$s = \frac{a*(1 - r^{n+1})}{1-r}$

This gives us the formula for an finite geometric series.

Interactive Demonstration

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